Problem: Simplify; express your answer in exponential form. Assume $r\neq 0, t\neq 0$. $\dfrac{{(r^{3})^{5}}}{{(r^{-5}t^{4})^{-1}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{3}}$ to the exponent ${5}$ . Now ${3 \times 5 = 15}$ , so ${(r^{3})^{5} = r^{15}}$ In the denominator, we can use the distributive property of exponents. ${(r^{-5}t^{4})^{-1} = (r^{-5})^{-1}(t^{4})^{-1}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{3})^{5}}}{{(r^{-5}t^{4})^{-1}}} = \dfrac{{r^{15}}}{{r^{5}t^{-4}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{15}}}{{r^{5}t^{-4}}} = \dfrac{{r^{15}}}{{r^{5}}} \cdot \dfrac{{1}}{{t^{-4}}} = r^{{15} - {5}} \cdot t^{- {(-4)}} = r^{10}t^{4}$.